The de-Broglie wavelength of a neutron at 27 ∘ C is λ . What will be its wavelength at 927 ∘ C Options: A . &nbspλ2 B . &nbspλ3 C . &nbspλ4 D . &nbspλ9

Answer: Option A : A 1 2 m V 2 = 3 2 k T ⇒ m V = √ 3 m K T λ = h p = h m V = h √ 3 m K T ⇒ λ α = 1 √ T ⇒ λ 1 λ 2 = √ T 2 T 1 = √ 273 + 927 273 + 27 = 2 ⇒ λ 2 = λ 1 2 = λ 2 .

The de-Broglie wavelength of a neutron at 27∘C is λ. What will be i

Question

The de-Broglie wavelength of a neutron at

27^{\circ} C

λ

. What will be its wavelength at

927^{\circ} C

Options:

A . λ2

B . λ3

C . λ4

D . λ9

Answer: Option A
:
A

\frac{1}{2} m V^{2} = \frac{3}{2} k T

\Rightarrow m V = \sqrt{3 m K T}

λ = \frac{h}{p} = \frac{h}{m V} = \frac{h}{\sqrt{3 m K T}}

\Rightarrow λ α = \frac{1}{\sqrt{T}}

\Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{273 + 927}{273 + 27}} = 2

\Rightarrow λ_{2} = \frac{λ_{1}}{2} = \frac{λ}{2} .

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